Question: $ g(x) = \int_{-8}^{x}(-2t + 4)\,dt\,$ $ g\,'(-3)\, =$
The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = -2t +4$ is continuous on $[-8,-3]$. Applying the theorem We're given: $ g(x) = \int_{-8}^{x}(-2t + 4)\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =-2x + 4$ Evaluating $g'(-3)$ $ g'(-3)= -2(-3) + 4 = 10$ The answer: $g'(-3)=10$